Let $X$ and $Y$ be independent random variables. Prove that $$P(X \leq x\mid X >Y) \leq P(X \leq x).$$
I try to use the definition of conditional probability, but I can't get success. $$P(X \leq x\mid X >Y) = \frac{P(X \leq x, X > Y)}{P(X >Y)} = \frac{P(Y < X \leq x)}{P(X >Y)}$$
Any ideas?
We have $F(x) F(y) \leq F(y)$. Hence, $F(x)-F(y) \leq F(x) [1-F(y)]$. In other words, $P\{y<X \leq x\} \leq P\{X \leq x\} P\{y<X\}$. If you integrate this with respect to the distribution of $Y$ (and use independence) you get $P\{Y<X \leq x\} \leq P\{X \leq x\} P\{Y<X\}$ which is the given inequality.